JEE Advanced 2025 Paper 2 · Q11 · Probability
A factory has three manufacturing units $M_1, M_2, M_3$ producing bulbs independently in proportions $2:2:1$. $20\%$ of all bulbs are defective. Of bulbs produced by $M_1$, $15\%$ are defective. Given that a randomly chosen defective bulb came from $M_2$ with probability $\frac{2}{5}$, find the probability that a bulb chosen from $M_3$ is defective.
Reveal answer + step-by-step solution
Correct answer:0.3
Solution
Total proportions: $P(M_1) = 2/5$, $P(M_2) = 2/5$, $P(M_3) = 1/5$. Let $p_i = P(\text{defective} \mid M_i)$. Bayes for $M_2$: $P(M_2 \mid \text{def}) = \frac{P(M_2) \cdot p_2}{P(\text{def})} = \frac{(2/5) p_2}{0.20} = \frac{2}{5}$ $\to$ $(2/5) p_2 = 0.20 \cdot (2/5) = 0.08$ $\to$ $p_2 = 0.20$. Total defective probability: $0.20 = (2/5)(0.15) + (2/5)(0.20) + (1/5) p_3 = 0.06 + 0.08 + 0.2 p_3$ $\to$ $0.2 p_3 = 0.06$ $\to$ $p_3 = 0.30$.
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